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I've been working in the following problem but I can't fill in all the details.

Consider $E=\mathcal{C}\big([0, 1],\mathbb C \big)$ endowed with the norm induced by the product $\displaystyle\left\langle f,g\right\rangle =\int_0^1 f(t)\overline{g(t)}\,dt$. Fix $\alpha\in (0, 1)$ and define $u_\alpha\,\colon E\to \mathbb C$ by $$\displaystyle u_\alpha(f):= \int_0^\alpha t^2 f(t)\,dt.$$ I must prove that $u_\alpha\in E^*$, and also have to find its norm. Furthermore, I have to explain why $E$ is not a Hilbert space using this operator.

It is easy to see that $u_\alpha$ is a continuous linear functional $E\to \mathbb C$ because $|\phi_\alpha(f)|\leq \sqrt{\dfrac{\alpha^5}{5}}\, \lVert f \rVert_E$, and thus $ \lVert u_\alpha \rVert \leq\sqrt{\dfrac{\alpha^5}{5}}$, and I conjectured that this must be its norm, but I have failed trying to prove that (it may not be its norm in the first place, but I haven't got anything better)

For the second item, it suffices to show that the Riesz-Fréchet representation theorem doesn't hold for $u_\alpha$ (I've done it already, it is not hard at all though it is not evident)

One last question, for $n\in \mathbb N$, what is the norm of the following operator: $\displaystyle \phi_\alpha(f)= \int_0^\alpha t^n f(t)\,dt $?

Can you help me to find such norms? Thanks in advance!

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  • $\begingroup$ How did you get the square root? (of $\alpha^5/5$) $\endgroup$ – Quoka Mar 6 '18 at 23:32
  • $\begingroup$ I just used the Cauchy-Schwarz inequality $\endgroup$ – EternalBlood Mar 6 '18 at 23:36
  • $\begingroup$ Using Cauchy-Scwartz I get $\alpha^5/5$, without the square root $\endgroup$ – Quoka Mar 6 '18 at 23:39
  • $\begingroup$ This is probably what you did: ${\displaystyle \left|\int _{\Omega}f(x){\overline {g(x)}}\,dx\right|^{2}\leq \int _{\Omega}|f(x)|^{2}\,dx\cdot \int _{\Omega}|g(x)|^{2}\,dx.}$ But one must extract square root to both sides $\endgroup$ – EternalBlood Mar 6 '18 at 23:59
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    $\begingroup$ Ok so to see that $\sqrt{\alpha^5/5}$ is actually the norm consider $f(t) = t^2\chi_{(0,a)}\sqrt{5/\alpha^5}$ $\endgroup$ – Quoka Mar 7 '18 at 0:09
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The Cauchy-Schwarz inequality becomes equality iff one of the vectors is a multiple of the other, so take $f(t) = \sqrt{\frac{5}{\alpha^5}} t^2$. Then note that $\|f \|_2 = 1$ and $$\lvert u_\alpha(f) \rvert= \sqrt{\frac{5}{\alpha^5}}\int^\alpha_0 t^4 dt = \sqrt{\frac{\alpha^5}{5}}, $$ thus the equality is achieved and so $\|u_\alpha\| = \sqrt{\frac{\alpha^5}{5}}$.

The exact same reasoning works for $$\phi_\alpha(f) = \int^\alpha_0 t^n f(t) dt;$$ just use Cauchy-Schwarz and then choose $f$ so as to attain equality.

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